How to count string occurrence in string?

Question

How can I count the number of times a particular string occurs in another string. For example, this is what I am trying to do in Javascript:

var temp = "This is a string.";
alert(temp.count("is")); //should output '2'

Answer 1

The g in the regular expression (short for global) says to search the whole string rather than just find the first occurrence. This matches is twice:

var temp = "This is a string.";
var count = (temp.match(/is/g) || []).length;
console.log(count);

And, if there are no matches, it returns 0:

var temp = "Hello World!";
var count = (temp.match(/is/g) || []).length;
console.log(count);

Answer 2

/** Function that count occurrences of a substring in a string;
 * @param {String} string               The string
 * @param {String} subString            The sub string to search for
 * @param {Boolean} [allowOverlapping]  Optional. (Default:false)
 *
 * @author Vitim.us https://gist.github.com/victornpb/7736865
 * @see Unit Test https://jsfiddle.net/Victornpb/5axuh96u/
 * @see http://stackoverflow.com/questions/4009756/how-to-count-string-occurrence-in-string/7924240#7924240
 */
function occurrences(string, subString, allowOverlapping) {

    string += "";
    subString += "";
    if (subString.length <= 0) return (string.length + 1);

    var n = 0,
        pos = 0,
        step = allowOverlapping ? 1 : subString.length;

    while (true) {
        pos = string.indexOf(subString, pos);
        if (pos >= 0) {
            ++n;
            pos += step;
        } else break;
    }
    return n;
}

Usage

occurrences("foofoofoo", "bar"); //0

occurrences("foofoofoo", "foo"); //3

occurrences("foofoofoo", "foofoo"); //1

allowOverlapping

occurrences("foofoofoo", "foofoo", true); //2

Matches:

  foofoofoo
1 `----´
2    `----´

Unit Test

• https://jsfiddle.net/Victornpb/5axuh96u/

Benchmark

I've made a benchmark test and my function is more then 10 times faster then the regexp match function posted by gumbo. In my test string is 25 chars length. with 2 occurences of the character 'o'. I executed 1 000 000 times in Safari.

Safari 5.1

Benchmark> Total time execution: 5617 ms (regexp)

Benchmark> Total time execution: 881 ms (my function 6.4x faster)

Firefox 4

Benchmark> Total time execution: 8547 ms (Rexexp)

Benchmark> Total time execution: 634 ms (my function 13.5x faster)

---

Edit: changes I've made

• cached substring length

• added type-casting to string.

• added optional 'allowOverlapping' parameter

• fixed correct output for "" empty substring case.

• https://gist.github.com/victornpb/7736865

Answer 3

function countInstances(string, word) {
   return string.split(word).length - 1;
}

Answer 4

You can try this:

var theString = "This is a string.";
console.log(theString.split("is").length - 1);

Answer 5

My solution:

var temp = "This is a string.";

function countOcurrences(str, value) {
  var regExp = new RegExp(value, "gi");
  return (str.match(regExp) || []).length;
}

console.log(countOcurrences(temp, 'is'));

Answer 6

You can use match to define such function:

String.prototype.count = function(search) {
    var m = this.match(new RegExp(search.toString().replace(/(?=[.\\+*?[^\]$(){}\|])/g, "\\"), "g"));
    return m ? m.length:0;
}

Answer 7

The non-regex version:

 var string = 'This is a string',
    searchFor = 'is',
    count = 0,
    pos = string.indexOf(searchFor);

while (pos > -1) {
    ++count;
    pos = string.indexOf(searchFor, ++pos);
}

console.log(count);   // 2

Answer 8

Just code-golfing Rebecca Chernoff's solution :-)

alert(("This is a string.".match(/is/g) || []).length);

Answer 9

String.prototype.Count = function (find) {
    return this.split(find).length - 1;
}

console.log("This is a string.".Count("is"));

This will return 2.

Answer 10

Here is the fastest function!

Why is it faster?

• Doesn't check char by char (with 1 exception)
• Uses a while and increments 1 var (the char count var) vs. a for loop checking the length and incrementing 2 vars (usually var i and a var with the char count)
• Uses WAY less vars
• Doesn't use regex!
• Uses an (hopefully) highly optimized function

• All operations are as combined as they can be, avoiding slowdowns due to multiple operations

  String.prototype.timesCharExist=function(c){var t=0,l=0,c=(c+'')[0];while(l=this.indexOf(c,l)+1)++t;return t};
  

Here is a slower and more readable version:

    String.prototype.timesCharExist = function ( chr ) {
        var total = 0, last_location = 0, single_char = ( chr + '' )[0];
        while( last_location = this.indexOf( single_char, last_location ) + 1 )
        {
            total = total + 1;
        }
        return total;
    };

This one is slower because of the counter, long var names and misuse of 1 var.

To use it, you simply do this:

    'The char "a" only shows up twice'.timesCharExist('a');

Edit: (2013/12/16)

DON'T use with Opera 12.16 or older! it will take almost 2.5x more than the regex solution!

On chrome, this solution will take between 14ms and 20ms for 1,000,000 characters.

The regex solution takes 11-14ms for the same amount.

Using a function (outside String.prototype) will take about 10-13ms.

Here is the code used:

    String.prototype.timesCharExist=function(c){var t=0,l=0,c=(c+'')[0];while(l=this.indexOf(c,l)+1)++t;return t};

    var x=Array(100001).join('1234567890');

    console.time('proto');x.timesCharExist('1');console.timeEnd('proto');

    console.time('regex');x.match(/1/g).length;console.timeEnd('regex');

    var timesCharExist=function(x,c){var t=0,l=0,c=(c+'')[0];while(l=x.indexOf(c,l)+1)++t;return t;};

    console.time('func');timesCharExist(x,'1');console.timeEnd('func');

The result of all the solutions should be 100,000!

Note: if you want this function to count more than 1 char, change where is c=(c+'')[0] into c=c+''

Answer 11

var temp = "This is a string.";
console.log((temp.match(new RegExp("is", "g")) || []).length);

Answer 12

I think the purpose for regex is much different from indexOf. indexOf simply find the occurance of a certain string while in regex you can use wildcards like [A-Z] which means it will find any capital character in the word without stating the actual character.

Example:

 var index = "This is a string".indexOf("is");
 console.log(index);
 var length = "This is a string".match(/[a-z]/g).length;
 // where [a-z] is a regex wildcard expression thats why its slower
 console.log(length);

Answer 13

Super duper old, but I needed to do something like this today and only thought to check SO afterwards. Works pretty fast for me.

String.prototype.count = function(substr,start,overlap) {
    overlap = overlap || false;
    start = start || 0;

    var count = 0, 
        offset = overlap ? 1 : substr.length;

    while((start = this.indexOf(substr, start) + offset) !== (offset - 1))
        ++count;
    return count;
};

Answer 14

       var myString = "This is a string.";
        var foundAtPosition = 0;
        var Count = 0;
        while (foundAtPosition != -1)
        {
            foundAtPosition = myString.indexOf("is",foundAtPosition);
            if (foundAtPosition != -1)
            {
                Count++;
                foundAtPosition++;
            }
        }
        document.write("There are " + Count + " occurrences of the word IS");

Refer :- count a substring appears in the string for step by step explanation.

Answer 15

Building upon @Vittim.us answer above. I like the control his method gives me, making it easy to extend, but I needed to add case insensitivity and limit matches to whole words with support for punctuation. (e.g. "bath" is in "take a bath." but not "bathing")

The punctuation regex came from: https://stackoverflow.com/a/25575009/497745 (How can I strip all punctuation from a string in JavaScript using regex?)

function keywordOccurrences(string, subString, allowOverlapping, caseInsensitive, wholeWord)
{

    string += "";
    subString += "";
    if (subString.length <= 0) return (string.length + 1); //deal with empty strings

    if(caseInsensitive)
    {            
        string = string.toLowerCase();
        subString = subString.toLowerCase();
    }

    var n = 0,
        pos = 0,
        step = allowOverlapping ? 1 : subString.length,
        stringLength = string.length,
        subStringLength = subString.length;

    while (true)
    {
        pos = string.indexOf(subString, pos);
        if (pos >= 0)
        {
            var matchPos = pos;
            pos += step; //slide forward the position pointer no matter what

            if(wholeWord) //only whole word matches are desired
            {
                if(matchPos > 0) //if the string is not at the very beginning we need to check if the previous character is whitespace
                {                        
                    if(!/[\s\u2000-\u206F\u2E00-\u2E7F\\'!"#$%&\(\)*+,\-.\/:;<=>?@\[\]^_`{|}~]/.test(string[matchPos - 1])) //ignore punctuation
                    {
                        continue; //then this is not a match
                    }
                }

                var matchEnd = matchPos + subStringLength;
                if(matchEnd < stringLength - 1)
                {                        
                    if (!/[\s\u2000-\u206F\u2E00-\u2E7F\\'!"#$%&\(\)*+,\-.\/:;<=>?@\[\]^_`{|}~]/.test(string[matchEnd])) //ignore punctuation
                    {
                        continue; //then this is not a match
                    }
                }
            }

            ++n;                
        } else break;
    }
    return n;
}

Please feel free to modify and refactor this answer if you spot bugs or improvements.

Answer 16

For anyone that finds this thread in the future, note that the accepted answer will not always return the correct value if you generalize it, since it will choke on regex operators like $ and .. Here's a better version, that can handle any needle:

function occurrences (haystack, needle) {
  var _needle = needle
    .replace(/\[/g, '\\[')
    .replace(/\]/g, '\\]')
  return (
    haystack.match(new RegExp('[' + _needle + ']', 'g')) || []
  ).length
}

Answer 17

Try it

<?php 
$str = "33,33,56,89,56,56";
echo substr_count($str, '56');
?>

<script type="text/javascript">
var temp = "33,33,56,89,56,56";
var count = temp.match(/56/g);  
alert(count.length);
</script>

Answer 18

Simple version without regex:

var temp = "This is a string.";

var count = (temp.split('is').length - 1);

alert(count);

Answer 19

No one will ever see this, but it's good to bring back recursion and arrow functions once in a while (pun gloriously intended)

String.prototype.occurrencesOf = function(s, i) {
 return (n => (n === -1) ? 0 : 1 + this.occurrencesOf(s, n + 1))(this.indexOf(s, (i || 0)));
};

Answer 20

Now this is a very old thread i've come across but as many have pushed their answer's, here is mine in a hope to help someone with this simple code.

var search_value = "This is a dummy sentence!";
var letter = 'a'; /*Can take any letter, have put in a var if anyone wants to use this variable dynamically*/
letter = letter && "string" === typeof letter ? letter : "";
var count;
for (var i = count = 0; i < search_value.length; count += (search_value[i++] == letter));
console.log(count);

I'm not sure if it is the fastest solution but i preferred it for simplicity and for not using regex (i just don't like using them!)

Answer 21

 function substrCount( str, x ) {
   let count = -1, pos = 0;
   do {
     pos = str.indexOf( x, pos ) + 1;
     count++;
   } while( pos > 0 );
   return count;
 }

Answer 22

ES2020 offers a new MatchAll which might be of use in this particular context.

Here we create a new RegExp, please ensure you pass 'g' into the function.

Convert the result using Array.from and count the length, which returns 2 as per the original requestor's desired output.

let strToCheck = RegExp('is', 'g')
let matchesReg = "This is a string.".matchAll(strToCheck)
console.log(Array.from(matchesReg).length) // 2

Answer 23

A simple way would be to split the string on the required word, the word for which we want to calculate the number of occurences, and subtract 1 from the number of parts:

function checkOccurences(string, word) {
      return string.split(word).length - 1;
}
const text="Let us see. see above, see below, see forward, see backward, see left, see right until we will be right"; 
const count=countOccurences(text,"see "); // 2

Answer 24

var countInstances = function(body, target) {
  var globalcounter = 0;
  var concatstring  = '';
  for(var i=0,j=target.length;i<body.length;i++){
    concatstring = body.substring(i-1,j);
    
    if(concatstring === target){
       globalcounter += 1;
       concatstring = '';
    }
  }
  
  
  return globalcounter;
 
};

console.log(   countInstances('abcabc', 'abc')   ); // ==> 2
console.log(   countInstances('ababa', 'aba')   ); // ==> 2
console.log(   countInstances('aaabbb', 'ab')   ); // ==> 1

Answer 25

---

substr_count translated to Javascript from php

---

• Locutus (Package that translates Php to JS)
• substr_count (official page, code copied below)

function substr_count (haystack, needle, offset, length) { 
  // eslint-disable-line camelcase
  //  discuss at: https://locutus.io/php/substr_count/
  // original by: Kevin van Zonneveld (https://kvz.io)
  // bugfixed by: Onno Marsman (https://twitter.com/onnomarsman)
  // improved by: Brett Zamir (https://brett-zamir.me)
  // improved by: Thomas
  //   example 1: substr_count('Kevin van Zonneveld', 'e')
  //   returns 1: 3
  //   example 2: substr_count('Kevin van Zonneveld', 'K', 1)
  //   returns 2: 0
  //   example 3: substr_count('Kevin van Zonneveld', 'Z', 0, 10)
  //   returns 3: false

  var cnt = 0

  haystack += ''
  needle += ''
  if (isNaN(offset)) {
    offset = 0
  }
  if (isNaN(length)) {
    length = 0
  }
  if (needle.length === 0) {
    return false
  }
  offset--

  while ((offset = haystack.indexOf(needle, offset + 1)) !== -1) {
    if (length > 0 && (offset + needle.length) > length) {
      return false
    }
    cnt++
  }

  return cnt
}

Check out Locutus's Translation Of Php's substr_count function

Answer 26

You could try this

let count = s.length - s.replace(/is/g, "").length;

Answer 27

The parameters: ustring: the superset string countChar: the substring

A function to count substring occurrence in JavaScript:

function subStringCount(ustring, countChar){
  var correspCount = 0;
  var corresp = false;
  var amount = 0;
  var prevChar = null;
  
 for(var i=0; i!=ustring.length; i++){

     if(ustring.charAt(i) == countChar.charAt(0) && corresp == false){
       corresp = true;
       correspCount += 1;
       if(correspCount == countChar.length){
         amount+=1;
         corresp = false;
         correspCount = 0;
       }
       prevChar = 1;
     }
     else if(ustring.charAt(i) == countChar.charAt(prevChar) && corresp == true){
       correspCount += 1;
       if(correspCount == countChar.length){
         amount+=1;
         corresp = false;
         correspCount = 0;
         prevChar = null;
       }else{
         prevChar += 1 ;
       }
     }else{
       corresp = false;
       correspCount = 0;
     }
 } 
 return amount;
}

console.log(subStringCount('Hello World, Hello World', 'll'));

Answer 28

var str = 'stackoverflow';
var arr = Array.from(str);
console.log(arr);

for (let a = 0; a <= arr.length; a++) {
  var temp = arr[a];
  var c = 0;
  for (let b = 0; b <= arr.length; b++) {
    if (temp === arr[b]) {
      c++;
    }

  }
  console.log(`the ${arr[a]} is counted for ${c}`)
}

Answer 29

Iterate less the second time (just when first letter of substring matches) but still uses 2 for loops:

   function findSubstringOccurrences(str, word) {
        let occurrences = 0;
        for(let i=0; i<str.length; i++){
            if(word[0] === str[i]){ // to make it faster and iterate less
                for(let j=0; j<word.length; j++){
                    if(str[i+j] !== word[j]) break;
                    if(j === word.length - 1) occurrences++;
                }
            }
        }
        return occurrences;
    }
    
    console.log(findSubstringOccurrences("jdlfkfomgkdjfomglo", "omg"));

Answer 30

//Try this code

const countSubStr = (str, search) => {
    let arrStr = str.split('');
    let i = 0, count = 0;

    while(i < arrStr.length){
        let subStr = i + search.length + 1 <= arrStr.length ?
                  arrStr.slice(i, i+search.length).join('') :
                  arrStr.slice(i).join('');
        if(subStr === search){
            count++;
            arrStr.splice(i, search.length);
        }else{
            i++;
        }
    }
    return count;
  }